Answer:
[tex]5.3\times 10^3 N/C[/tex]
Explanation:
We are given that
Distance between plates=d=2.2 cm=[tex]2.2\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
[tex]\sigma=47nC/m^2=47\times 10^{-9}C/m^2[/tex]
Using [tex]1 nC=10^{-9} C[/tex]
We have to find the magnitude of E in the region between the plates.
We know that the electric field for parallel plates
[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
[tex]E_1=\frac{\sigma}{2\epsilon_0}[/tex]
[tex]E_2=\frac{\sigma}{2\epsilon_0}[/tex]
[tex]E=E_1+E_2[/tex]
[tex]E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]
Substitute the values
[tex]E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}[/tex]
[tex]E=5.3\times 10^3 N/C[/tex]
Hence, the magnitude of E in the region between the plates=[tex]5.3\times 10^3 N/C[/tex]
